Question: $\lim_{x\to\infty}\dfrac{e^{5x-3}}{\ln(x-2)}=?$ Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $5$ (Choice C) C $e^5$ (Choice D) D $\infty$
Answer: $\lim_{x\to\infty} e^{5x-3}=\infty$ and $\lim_{x\to\infty} \ln(x-2)=\infty$, so $\lim_{x\to\infty}\dfrac{e^{5x-3}}{\ln(x-2)}$ results in the indeterminate form $\dfrac{\infty}{\infty}$. We should use l'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to\infty}\dfrac{e^{5x-3}}{\ln(x-2)} \\\\ &=\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[e^{5x-3}\right]}{\dfrac{d}{dx}[\ln(x-2)]} \gray{\text{l'Hôpital's rule}} \\\\ &=\lim_{x\to\infty}\dfrac{5e^{5x-3}}{\left(\dfrac{1}{x-2}\right)} \\\\ &=\lim_{x\to\infty}5(x-2)e^{5x-3} \\\\ &=\infty \end{aligned}$ Note that we were only able to use l'Hôpital's rule because the limit $\lim_{x\to\infty}\dfrac{\dfrac{d}{dx}\left[e^{5x-3}\right]}{\dfrac{d}{dx}[\ln(x-2)]}$ can actually be determined. In conclusion, $\lim_{x\to\infty}\dfrac{e^{5x-3}}{\ln(x-2)}=\infty$